Problem: The equation of a circle $C$ is $x^2+y^2+14x-10y+25 = 0$. What is its center $(h, k)$ and its radius $r$ ?
Solution: To find the equation in standard form, complete the square. $(x^2+14x) + (y^2-10y) = -25$ $(x^2+14x+49) + (y^2-10y+25) = -25 + 49 + 25$ $(x+7)^{2} + (y-5)^{2} = 49 = 7^2$ Thus, $(h, k) = (-7, 5)$ and $r = 7$.